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MCQOPTIONS
This section includes 12583 Mcqs, each offering curated multiple-choice questions to sharpen your Joint Entrance Exam - Main (JEE Main) knowledge and support exam preparation. Choose a topic below to get started.
| 5701. |
A cell of e.m.f. \[E\] is connected with an external resistance \[R\], then p.d. across cell is \[V\]. The internal resistance of cell will be [MNR 1987; Kerala PMT 2002; MP PMT 2002] |
| A. | \[\frac{(E-V)R}{E}\] |
| B. | \[\frac{(E-V)R}{V}\] |
| C. | \[\frac{(V-E)R}{V}\] |
| D. | \[\frac{(V-E)R}{E}\] |
| Answer» C. \[\frac{(V-E)R}{V}\] | |
| 5702. |
A current of 2.0 ampere passes through a cell of e.m.f. 1.5 volts having internal resistance of 0.15 ohm. The potential difference measured, in volts, across both the ends of the cell will be [UPSEAT 1999, 2000] |
| A. | 1.35 |
| B. | 1.50 |
| C. | 1.00 |
| D. | 1.20 |
| Answer» E. | |
| 5703. |
The current in the given circuit is [AIIMS 2000; MH CET 2003] |
| A. | 0.1 A |
| B. | 0.2 A |
| C. | 0.3 A |
| D. | 0.4 A |
| Answer» B. 0.2 A | |
| 5704. |
The internal resistance of a cell of e.m.f. 12V is \[5\times {{10}^{-2}}\,\Omega \]. It is connected across an unknown resistance. Voltage across the cell, when a current of 60 A is drawn from it, is [CBSE PMT 2000] |
| A. | 15 V |
| B. | 12 V |
| C. | 9 V |
| D. | 6 V |
| Answer» D. 6 V | |
| 5705. |
In the given circuit the current I1 is [DCE 2000] |
| A. | 0.4 A |
| B. | ? 0.4 A |
| C. | 0.8 A |
| D. | ? 0.8 A |
| Answer» C. 0.8 A | |
| 5706. |
A battery having e.m.f. \[5\,\,V\] and internal resistance 0.5 W is connected with a resistance of 4.5 W then the voltage at the terminals of battery is [RPMT 2000] |
| A. | 4.5 V |
| B. | 4 V |
| C. | 0 V |
| D. | 2 V |
| Answer» B. 4 V | |
| 5707. |
A storage cell is charged by 5 amp D.C. for 18 hours. Its strength after charging will be [JIPMER 1999] |
| A. | 18 AH |
| B. | 5 AH |
| C. | 90 AH |
| D. | 15 AH |
| Answer» D. 15 AH | |
| 5708. |
A cell of emf 6 V and resistance 0.5 ohm is short circuited. The current in the cell is [JIPMER 1999] |
| A. | 3 amp |
| B. | 12 amp |
| C. | 24 amp |
| D. | 6 amp |
| Answer» C. 24 amp | |
| 5709. |
Electromotive force is the force, which is able to maintain a constant [Pb. PMT 1999] |
| A. | Current |
| B. | Resistance |
| C. | Power |
| D. | Potential difference |
| Answer» E. | |
| 5710. |
Four identical cells each having an electromotive force (e.m.f.) of 12V, are connected in parallel. The resultant electromotive force (e.m.f.) of the combination is [CPMT 1999] |
| A. | 48 V |
| B. | 12 V |
| C. | 4 V |
| D. | 3 V |
| Answer» C. 4 V | |
| 5711. |
In the shown circuit, what is the potential difference across A and B [AIIMS 1999] |
| A. | 50 V |
| B. | 45 V |
| C. | 30 V |
| D. | 20 V |
| Answer» E. | |
| 5712. |
The e.m.f. of a cell is E volts and internal resistance is \[r\] ohm. The resistance in external circuit is also\[r\] ohm. The p.d. across the cell will be [CPMT 1985; NCERT 1973] |
| A. | E/2 |
| B. | 2E |
| C. | 4E |
| D. | E/4 |
| Answer» B. 2E | |
| 5713. |
Two batteries of e.m.f. 4V and 8 V with internal resistances 1 W and 2 W are connected in a circuit with a resistance of 9 W as shown in figure. The current and potential difference between the points P and Q are [AFMC 1999] |
| A. | \[\frac{1}{3}A\,\,\text{and }\,3V\] |
| B. | \[\frac{1}{6}A\,\,\text{and}\,\text{ }4V\] |
| C. | \[\frac{1}{9}A\,\,\text{and}\,\text{ 9}V\] |
| D. | \[\frac{1}{2}A\,\,\text{and}\,\,12V\] |
| Answer» B. \[\frac{1}{6}A\,\,\text{and}\,\text{ }4V\] | |
| 5714. |
For driving a current of 2 A for 6 minutes in a circuit, 1000 J of work is to be done. The e.m.f. of the source in the circuit is [CPMT 1999] |
| A. | 1.38 V |
| B. | 1.68 V |
| C. | 2.04 V |
| D. | 3.10 V |
| Answer» B. 1.68 V | |
| 5715. |
Emf is most closely related to [DCE 1999] |
| A. | Mechanical force |
| B. | Potential difference |
| C. | Electric field |
| D. | Magnetic field |
| Answer» C. Electric field | |
| 5716. |
The number of dry cells, each of e.m.f. 1.5 volt and internal resistance 0.5 ohm that must be joined in series with a resistance of 20 ohm so as to send a current of 0.6 ampere through the circuit is [SCRA 1998] |
| A. | 2 |
| B. | 8 |
| C. | 10 |
| D. | 12 |
| Answer» D. 12 | |
| 5717. |
A storage battery has e.m.f. 15 volts and internal resistance 0.05 ohm. Its terminal voltage when it is delivering 10 ampere is [JIPMER 1997] |
| A. | 30 volts |
| B. | 1.00 volts |
| C. | 14.5 volts |
| D. | 15.5 volts |
| Answer» D. 15.5 volts | |
| 5718. |
Two resistances \[{{R}_{1}}\] and \[{{R}_{2}}\] are joined as shown in the figure to two batteries of e.m.f. \[{{E}_{1}}\] and \[{{E}_{2}}\]. If \[{{E}_{2}}\] is short-circuited, the current through \[{{R}_{1}}\] is [NDA 1995] |
| A. | \[{{E}_{1}}/{{R}_{1}}\] |
| B. | \[{{E}_{2}}/{{R}_{1}}\] |
| C. | \[{{E}_{2}}/{{R}_{2}}\] |
| D. | \[{{E}_{1}}/({{R}_{2}}+{{R}_{1}})\] |
| Answer» B. \[{{E}_{2}}/{{R}_{1}}\] | |
| 5719. |
If \[{{V}_{AB}}=4V\] in the given figure, then resistance X will be [RPET 1997] |
| A. | \[5\,\Omega \] |
| B. | \[10\,\Omega \] |
| C. | \[15\,\Omega \] |
| D. | \[20\,\Omega \] |
| Answer» E. | |
| 5720. |
Consider the circuit shown in the figure. The current \[{{I}_{3}}\] is equal to [AMU 1995] |
| A. | 5 amp |
| B. | 3 amp |
| C. | \[-3\,amp\] |
| D. | \[-5/6\,amp\] |
| Answer» E. | |
| 5721. |
If six identical cells each having an e.m.f. of 6V are connected in parallel, the e.m.f. of the combination is [EAMCET (Med.) 1995; Pb. PMT 1999; CPMT 2000] |
| A. | 1 V |
| B. | 36 V |
| C. | \[\frac{1}{6}V\] |
| D. | 6 V |
| Answer» E. | |
| 5722. |
Two non-ideal identical batteries are connected in parallel. Consider the following statements [MP PMT 1999] (i) The equivalent e.m.f. is smaller than either of the two e.m.f.s (ii) The equivalent internal resistance is smaller than either of the two internal resistances |
| A. | Both (i) and (ii) are correct |
| B. | (i) is correct but (ii) is wrong |
| C. | (ii) is correct but (i) is wrong |
| D. | Both (i) and (ii) are wrong |
| Answer» D. Both (i) and (ii) are wrong | |
| 5723. |
By a cell a current of 0.9 A flows through 2 ohm resistor and 0.3 A through 7 ohm resistor. The internal resistance of the cell is [KCET 2003] |
| A. | \[0.5\,\Omega \] |
| B. | \[1.0\,\Omega \] |
| C. | \[1.2\,\Omega \] |
| D. | \[2.0\,\Omega \] |
| Answer» B. \[1.0\,\Omega \] | |
| 5724. |
When a resistance of 2 ohm is connected across the terminals of a cell, the current is 0.5 A. When the resistance is increased to 5 ohm, the current is 0.25 A. The e.m.f. of the cell is [MP PET 1999, 2000; Pb. PMT 2002; MP PMT 2000] |
| A. | 1.0 V |
| B. | 1.5 V |
| C. | 2.0 V |
| D. | 2.5 V |
| Answer» C. 2.0 V | |
| 5725. |
The current in the arm CD of the circuit will be [MP PMT/PET 1998; MP PMT 2000; DPMT 2000] |
| A. | \[{{i}_{1}}+{{i}_{2}}\] |
| B. | \[{{i}_{2}}+{{i}_{3}}\] |
| C. | \[{{i}_{1}}+{{i}_{3}}\] |
| D. | \[{{i}_{1}}-{{i}_{2}}+{{i}_{3}}\] |
| Answer» C. \[{{i}_{1}}+{{i}_{3}}\] | |
| 5726. |
100 cells each of e.m.f. 5 V and internal resistance 1 ohm are to be arranged so as to produce maximum current in a 25 ohms resistance. Each row is to contain equal number of cells. The number of rows should be [MP PMT 1997] |
| A. | 2 |
| B. | 4 |
| C. | 5 |
| D. | 10 |
| Answer» B. 4 | |
| 5727. |
A current of two amperes is flowing through a cell of e.m.f. 5 volts and internal resistance 0.5 ohm from negative to positive electrode. If the potential of negative electrode is 10V, the potential of positive electrode will be [MP PMT 1997] |
| A. | 5 V |
| B. | 14 V |
| C. | 15 V |
| D. | 16 V |
| Answer» C. 15 V | |
| 5728. |
Consider the circuit given here with the following parameters E.M.F. of the cell = 12 V. Internal resistance of the cell \[=2\,\Omega \]. Resistance \[R=4\,\Omega \] Which one of the following statements in true |
| A. | Rate of energy loss in the source is = 8 W |
| B. | Rate of energy conversion in the source is 16 W |
| C. | Power output in is = 8 W |
| D. | Potential drop across R is = 16 V |
| Answer» B. Rate of energy conversion in the source is 16 W | |
| 5729. |
A dry cell has an e.m.f. of 1.5 V and an internal resistance of \[0.05\,\Omega \]. The maximum current obtainable from this cell for a very short time interval is [Haryana CEE 1996] |
| A. | 30 A |
| B. | 300 A |
| C. | 3 A |
| D. | 0.3 A |
| Answer» B. 300 A | |
| 5730. |
A battery of e.m.f. E and internal resistance r is connected to a variable resistor R as shown here. Which one of the following is true [MP PMT 1995] |
| A. | Potential difference across the terminals of the battery is maximum when R = r |
| B. | Power delivered to the resistor is maximum when R = r |
| C. | Current in the circuit is maximum when R = r |
| D. | Current in the circuit is maximum when \[R>>r\] |
| Answer» C. Current in the circuit is maximum when R = r | |
| 5731. |
The figure shows a network of currents. The magnitude of currents is shown here. The current i will be [MP PMT 1995] |
| A. | 3 A |
| B. | 13 A |
| C. | 23 A |
| D. | ? 3 A |
| Answer» D. ? 3 A | |
| 5732. |
The electromotive force of a primary cell is 2 volts. When it is short-circuited it gives a current of 4 amperes. Its internal resistance in ohms is [MP PET 1995] |
| A. | 0.5 |
| B. | 5.0 |
| C. | 2.0 |
| D. | 8.0 |
| Answer» B. 5.0 | |
| 5733. |
A torch battery consisting of two cells of 1.45 volts and an internal resistance \[0.15\,\Omega \], each cell sending currents through the filament of the lamps having resistance 1.5ohms. The value of current will be [MP PET 1994] |
| A. | 16.11 amp |
| B. | 1.611 amp |
| C. | 0.1611 amp |
| D. | 2.6 amp |
| Answer» C. 0.1611 amp | |
| 5734. |
A cell of \[e.m.f.\] \[1.5\,V\] having a finite internal resistance is connected to a load resistance of \[2\,\Omega \]. For maximum power transfer the internal resistance of the cell should be [BIT 1988] |
| A. | 4 ohm |
| B. | 0.5 ohm |
| C. | 2 ohm |
| D. | None of these |
| Answer» D. None of these | |
| 5735. |
The internal resistances of two cells shown are \[0.1\,\Omega \] and \[0.3\,\Omega \]. If \[R=0.2\,\Omega \], the potential difference across the cell |
| A. | B will be zero |
| B. | A will be zero |
| C. | A and B will be 2V |
| D. | A will be \[>2V\] and B will be \[<2V\] |
| Answer» B. A will be zero | |
| 5736. |
Two identical cells send the same current in \[2\,\Omega \] resistance, whether connected in series or in parallel. The internal resistance of the cell should be [NCERT 1982; Kerala PMT 2002] |
| A. | \[1\,\Omega \] |
| B. | \[2\,\Omega \] |
| C. | \[\frac{1}{2}\Omega \] |
| D. | \[2.5\,\Omega \] |
| Answer» C. \[\frac{1}{2}\Omega \] | |
| 5737. |
To get the maximum current from a parallel combination of n identical cells each of internal resistance r in an external resistance R, when [DPMT 1999] |
| A. | \[R>>r\] |
| B. | \[R<<r\] |
| C. | \[R=r\] |
| D. | None of these |
| Answer» C. \[R=r\] | |
| 5738. |
A cell of internal resistance r is connected to an external resistance R. The current will be maximum in R, if [CPMT 1982] |
| A. | \[R=r\] |
| B. | \[R<r\] |
| C. | \[R>r\] |
| D. | \[R=r/2\] |
| Answer» B. \[R<r\] | |
| 5739. |
n identical cells each of e.m.f. E and internal resistance r are connected in series. An external resistance R is connected in series to this combination. The current through R is [DPMT 2002] |
| A. | \[\frac{nE}{R+nr}\] |
| B. | \[\frac{nE}{nR+r}\] |
| C. | \[\frac{E}{R+nr}\] |
| D. | \[\frac{nE}{R+r}\] |
| Answer» B. \[\frac{nE}{nR+r}\] | |
| 5740. |
The internal resistance of a cell depends on |
| A. | The distance between the plates |
| B. | The area of the plates immersed |
| C. | The concentration of the electrolyte |
| D. | All the above |
| Answer» E. | |
| 5741. |
When cells are connected in parallel, then [MNR 1983] |
| A. | The current decreases |
| B. | The current increases |
| C. | The e.m.f. increases |
| D. | The e.m.f. decreases |
| Answer» C. The e.m.f. increases | |
| 5742. |
The reading of a high resistance voltmeter when a cell is connected across it is 2.2 V. When the terminals of the cell are also connected to a resistance of \[5\,\Omega \] the voltmeter reading drops to 1.8 V. Find the internal resistance of the cell [KCET 2003; MP PMT 2003] |
| A. | \[1.2\,\Omega \] |
| B. | \[1.3\,\Omega \] |
| C. | \[1.1\,\Omega \] |
| D. | \[1.4\,\Omega \] |
| Answer» D. \[1.4\,\Omega \] | |
| 5743. |
A cell whose e.m.f. is 2 V and internal resistance is \[0.1\,\Omega \], is connected with a resistance of\[3.9\,\Omega \]. The voltage across the cell terminal will be [CPMT 1990; MP PET 1993; CBSE PMT 1999; AFMC 1999; Pb. PMT 2000; AIIMS 2001] |
| A. | \[0.50\,V\] |
| B. | \[1.90\,V\] |
| C. | \[1.95\,V\] |
| D. | \[2.00\,V\] |
| Answer» D. \[2.00\,V\] | |
| 5744. |
A new flashlight cell of e.m.f. 1.5 volts gives a current of 15 amps, when connected directly to an ammeter of resistance \[0.04\,\Omega \]. The internal resistance of cell is [MP PET 1994] |
| A. | \[0.04\,\Omega \] |
| B. | \[0.06\,\Omega \] |
| C. | \[0.10\,\Omega \] |
| D. | \[10\,\Omega \] |
| Answer» C. \[0.10\,\Omega \] | |
| 5745. |
The magnitude and direction of the current in the circuit shown will be [CPMT 1986, 88] |
| A. | \[\frac{7}{3}\]A from a to b through e |
| B. | \[\frac{7}{3}\]A from b to a through e |
| C. | 1A from b to a through e |
| D. | 1A from a to b through e |
| Answer» E. | |
| 5746. |
The potential difference in open circuit for a cell is 2.2 volts. When a 4 ohm resistor is connected between its two electrodes the potential difference becomes 2 volts. The internal resistance of the cell will be [MP PMT 1984; SCRA 1994; CBSE PMT 2002] |
| A. | 1 ohm |
| B. | 0.2 ohm |
| C. | 2.5 ohm |
| D. | 0.4 ohm |
| Answer» E. | |
| 5747. |
A 50V battery is connected across a 10 ohm resistor. The current is 4.5 amperes. The internal resistance of the battery is [CPMT 1985; BHU 1997; Pb. PMT 2001] |
| A. | Zero |
| B. | 0.5 ohm |
| C. | 1.1 ohm |
| D. | 5.0 ohm |
| Answer» D. 5.0 ohm | |
| 5748. |
A primary cell has an e.m.f. of 1.5 volts, when short-circuited it gives a current of 3 amperes. The internal resistance of the cell is [CPMT 1976, 83] |
| A. | 4.5\[ohm\] |
| B. | 2\[ohm\] |
| C. | 0.5\[ohm\] |
| D. | 1/4.5\[ohm\] |
| Answer» D. 1/4.5\[ohm\] | |
| 5749. |
The terminal potential difference of a cell when short-circuited is (\[E\] = E.M.F. of the cell) |
| A. | \[E\] |
| B. | \[E/2\] |
| C. | Zero |
| D. | \[E/3\] |
| Answer» D. \[E/3\] | |
| 5750. |
When a resistance of 2ohm is connected across the terminals of a cell, the current is 0.5 amperes. When the resistance is increased to 5 ohm, the current is 0.25 amperes. The internal resistance of the cell is [MP PMT 1996] |
| A. | \[0.5\,ohm\] |
| B. | \[1.0\,ohm\] |
| C. | \[1.5\,ohm\] |
| D. | \[2.0\,ohm\] |
| Answer» E. | |